(A.) Two astronauts, each having a mass of75.8 kg, are connected by a 14.3m rope

Question

(A.) Two astronauts, each having a mass of75.8 kg, are connected by a 14.3m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.77 m/s. Calculate the magnitude of the initial angular momentum of the system by treating the astronauts as particles.
Answer in units of kgm2/s

(B.)Calculate the rotational energy of the system.
Answer in units of J

(C.) By pulling the rope, the astronauts shorten the distance between them to 4.91 m. What is the new angular velocity of the astronauts? Answer in units of rad/s

(D.) How much work is done by the astronauts in shortening the rope?
Answer in units of J

Explanation / Answer

v = 5.77m/s

R = 14.3/2 = 7.15

=> = v/R = 5.77/ 7.15 = 0.807 rad/sec

initial angular momentum each astronaut = mr2 = 75.8*7.152*0.807

total systems angular momentum = 2*75.8*7.152*0.807 = 6254.39 kgm2/s

Rotational energy = 2*mr2*2/2 = 75.8*14.32*0.4032 = 2517.4 J

Angular momentum remains constant

=>(r1)21 = (r2)22

=> 2 = 7.152x0.807/2.4552 = 6.845 rad/sec

Workdone = change in energy =

(mr22)2 -(mr22)1 = angular momentum * (2-1) = 6254.39*(6.845-0.807) = 37764.911 J

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